The perimeter of a sector of angle 90° of a circle with radius 14 cm is ...............

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(a)

50 cm

Area of the ring of external radius R and internal radius r is .................

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(a)

\(\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) \)

The boundary of a circle is called its .................

1

(a)

circumference

Angle described by the minute hand in one minute is ..................

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(a)

6°

Circumference of a circle bears a constant ratio with its ................

1

(a)

diameter

Area of a circle is the portion enclosed under perimeter.

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(a) False

(b) True

Perimeter of a semicircle of radius r is equal to \(\pi r\).

1

(a) True

(b) False

If circumferences of two circles are equal, then their areas must be equal.

1

(a) False

(b) True

The perimeter of a semicircle is \(\pi r+2r\).

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(a) False

(b) True

The area of a sector is always greater than the area of the corresponding segment.

1

(a) True

(b) False

A chord of a circle of radius 14 cm subtends a right angle at the centre.  What is the area of the mirror sector? \([\pi\ = {22\over 7}]\)

1

(a)

Area of the sector = \({\theta \pi r^2\over 360^O}1={90^o\over 360^o}\times \pi \times (14)^2\) 
= \({1\over 4}\times {22\over 7}\times 14\times 14\)
= \(154\ cm^2\)

A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel. (use \(\pi ={22\over 7}\))

1

(a)

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

 A pendulum swings through an angle of \(30^o\) and describes an arc 8.8 cm in length.  Find the length of pendulum. ( use \(\pi ={22\over 7}\))

1

(a)

Let length of the pendulum he I cm.
Length of the arc = 8.8 cm
⇒ \(\frac { \theta \pi i }{ 180 } =8.8cm\Rightarrow \frac { 30°\times \pi \times i }{ 180° } =8.8\)
⇒ \(\pi i=8.8\times 6\Rightarrow \frac { 22 }{ 7 } \times i=52.8cm\)
⇒ \(i=\frac { 52.8\times 7 }{ 22 } cm\quad =\quad 16.8\quad cm\)

An arc of a circle is of length \(5\pi\) cm and the sector it bounds has an area of \(20\pi\ cm^2\). Find the radius of the circle

1

(a)

length of arc AB = 5\(\pi\) cm
Let AOB = \(\theta\)
Now   i = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r
5\(\pi\) = \(\frac { \theta }{ 180° } \)\(\pi\)r ⇒ \({900\over r}=\theta\)
Now, Area of sector = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r²
20\(\pi\) =  \(\frac { r }{ 360° } \)x 2\(\pi\)r² ⇒ 20 = \(900°\over360°\)r ⇒ r = 8 cm

The minute hand of a clock is \(\sqrt {21}\) cm long.  Find the area described by the minute hand on the face of the clock between 7.00 am and 7.05 am. \([Use\ \pi={22\over 7}]\)

1

(a)

Time taken by minute hand to make one circle = 60 minutes.
\(\therefore\) Angle described in 60 minutes = \(360^o\)
Angle described in 5 minutes [i.e., from 7.00 a.m. to 7.05 a.m.] = \({360^o\over 60^o}\times 5=30^o\)
Radius of circle = length of minute hand = \(\sqrt {21}cm\)
Area swept = \({\theta \over 360^o}\times r^2={30^oover 360^o}\times {22\over 7}\times \sqrt{21}\times \sqrt {21}={1\over 12}\times {22\over 7}\times 21 = {11\over 2}\ cm^2=5.5 cm^2\)